2011 AMC 10A Problem 25

Let $R$ be the unit square with corners $(0,0), (1,0), (1,1), (0,1)$, and let $X = (p, q)$ be interior. Each of the $n$ triangles has vertex $X$, and its opposite side lies along the boundary of $R$. A side of a triangle is a straight segment, so it cannot turn a corner of the square; therefore every corner must lie on one of the rays.

The four rays through the corners split $R$ into four triangles whose bases are the sides of the square: the triangle on the bottom side has area $\tfrac12 \cdot 1 \cdot q$, the one on the left has area $\tfrac{p}{2}$, and the others have areas $\tfrac{1-q}{2}$ and $\tfrac{1-p}{2}$. The remaining rays subdivide these. If the bottom triangle is cut into $k$ triangles of area $\tfrac1n$, then $\tfrac{q}{2} = \tfrac{k}{n}$, so $q = \tfrac{2k}{n}$. Likewise $1 - q$, $p$, $1 - p$ are positive multiples of $\tfrac{2}{n}$.

Conversely, if $p$ and $q$ are positive multiples of $\tfrac2n$ less than $1$, cutting each side into pieces of the right length gives triangles of area $\tfrac1n$ (same height, equal bases). So $X$ is $n$-ray partitional exactly when
$$p, q \in \left\{\tfrac{2}{n}, \tfrac{4}{n}, \dots, 1 - \tfrac{2}{n}\right\}.$$

For $n = 100$: $p, q \in \{\tfrac{1}{50}, \dots, \tfrac{49}{50}\}$, giving $49^2 = 2401$ points. For $n = 60$: $p, q \in \{\tfrac{1}{30}, \dots, \tfrac{29}{30}\}$. A coordinate belongs to both lists when $\tfrac{a}{50} = \tfrac{c}{30}$, i.e. $3a = 5c$, so $a$ is a multiple of $5$: $a = 5, 10, \dots, 45$, nine values (the multiples of $\tfrac1{10}$). Thus $9^2 = 81$ points are both.

The count is $2401 - 81 = 2320$.

The answer is $\boxed{\textbf{(C)}\ 2320}$.

The whole problem is the observation that the rays are forced through the corners; after that, "equal areas" for triangles sharing an apex and a base line is just "equal bases," and the corner-triangle areas depend only on the distances from $X$ to the sides. That converts a geometric condition into a grid of rational points, where the counting is routine and two grids meet on the coarser grid of step $\frac{1}{\gcd(50, 30)} = \frac{1}{10}$.

Sanity-check the structure with a small case: a point is $4$-ray partitional only if all four corner triangles have area $\tfrac14$, forcing $p = q = \tfrac12$, the center, which matches the formula ($1$ value per coordinate). For $n = 6$, $p, q \in \{\tfrac13, \tfrac23\}$: four points. Trusting the pattern, the answer is $(\tfrac{n}{2} - 1)^2$ for $n$ even, so $49^2 - 9^2 = (49 - 9)(49 + 9) = 40 \cdot 58 = 2320$.

• Counting 50 or 51 values per coordinate (including boundary points, which are not interior), or forgetting to subtract the points that are also 60-ray partitional.
• Miscounting the overlap: shared coordinates are multiples of $\tfrac{1}{10}$ strictly between $0$ and $1$, nine of them, not $10$ or $\gcd(49, 29)$.
• Allowing rays that miss the corners, which would make a region with a bent side; such a region is a quadrilateral, not a triangle.